Math Problem Statement

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If the perimeter of the Norman window is 23 ft, find the dimensions of r and w of the window that allow the window to admit the most light.

Solution

To solve this problem, we need to maximize the amount of light that the Norman window can admit, which involves maximizing the area of the window subject to the given constraint on the perimeter.

Step 1: Understand the geometry

The Norman window consists of:

  • A rectangular part of width ww and height hh.
  • A semicircular part with radius rr, where the diameter of the semicircle is equal to the width of the rectangle, meaning w=2rw = 2r.

Step 2: Express the perimeter

The perimeter consists of:

  • The perimeter of the semicircle: Semicircular perimeter=12×2πr=πr\text{Semicircular perimeter} = \frac{1}{2} \times 2\pi r = \pi r.
  • The perimeter of the rectangular sides (excluding the top, as it's already part of the semicircle): Rectangular perimeter=2h+w\text{Rectangular perimeter} = 2h + w, and since w=2rw = 2r, we can substitute this.

So, the total perimeter PP is given by: P=2h+2r+πrP = 2h + 2r + \pi r We are given that the total perimeter is 23 feet: 2h+2r+πr=232h + 2r + \pi r = 23 This is the constraint equation.

Step 3: Express the area

The area AA of the window is the sum of the area of the rectangle and the area of the semicircle: A=Rectangular area+Semicircular area=w×h+12πr2A = \text{Rectangular area} + \text{Semicircular area} = w \times h + \frac{1}{2}\pi r^2 Substitute w=2rw = 2r into the equation: A=2r×h+12πr2A = 2r \times h + \frac{1}{2} \pi r^2

Step 4: Solve the perimeter equation for hh

From the perimeter equation: 2h+(2+π)r=232h + (2 + \pi)r = 23 Solve for hh: 2h=23(2+π)r2h = 23 - (2 + \pi)r h=23(2+π)r2h = \frac{23 - (2 + \pi)r}{2}

Step 5: Substitute hh into the area equation

Substitute the expression for hh into the area equation: A=2r×23(2+π)r2+12πr2A = 2r \times \frac{23 - (2 + \pi)r}{2} + \frac{1}{2}\pi r^2 Simplify: A=r×(23(2+π)r)+12πr2A = r \times (23 - (2 + \pi)r) + \frac{1}{2}\pi r^2 A=23r(2+π)r2+12πr2A = 23r - (2 + \pi)r^2 + \frac{1}{2}\pi r^2 Combine terms: A=23r((2+π)12π)r2A = 23r - \left( (2 + \pi) - \frac{1}{2} \pi \right) r^2 A=23r(2+π2)r2A = 23r - \left( 2 + \frac{\pi}{2} \right) r^2

Step 6: Maximize the area

To find the value of rr that maximizes the area, we take the derivative of AA with respect to rr and set it equal to zero: dAdr=232×(2+π2)r=0\frac{dA}{dr} = 23 - 2 \times \left( 2 + \frac{\pi}{2} \right) r = 0 Solve for rr: 23=2×(2+π2)r23 = 2 \times \left( 2 + \frac{\pi}{2} \right) r r=232×(2+π2)r = \frac{23}{2 \times \left( 2 + \frac{\pi}{2} \right)} Simplify the denominator: r=232×(2+1.57)237.14r = \frac{23}{2 \times \left( 2 + 1.57 \right)} \approx \frac{23}{7.14} r3.22ftr \approx 3.22 \, \text{ft}

Step 7: Find ww and hh

Since w=2rw = 2r: w=2×3.226.44ftw = 2 \times 3.22 \approx 6.44 \, \text{ft}

Now, substitute rr back into the perimeter equation to find hh: 2h+2×3.22+π×3.22=232h + 2 \times 3.22 + \pi \times 3.22 = 23 2h+6.44+10.12232h + 6.44 + 10.12 \approx 23 2h+16.56232h + 16.56 \approx 23 2h6.442h \approx 6.44 h3.22fth \approx 3.22 \, \text{ft}

Final Answer:

  • r3.22ftr \approx 3.22 \, \text{ft}
  • w6.44ftw \approx 6.44 \, \text{ft}
  • h3.22fth \approx 3.22 \, \text{ft}

These dimensions maximize the amount of light the window can admit.

Do you need further clarification or details?

Here are 5 related questions:

  1. How does the radius affect the perimeter and area of the window?
  2. What would happen to the area if the perimeter constraint was increased?
  3. How does the presence of a semicircle alter the optimization compared to a simple rectangle?
  4. What are the units used for the area of the window?
  5. Can we solve this problem using a different method, such as Lagrange multipliers?

Tip: For optimization problems involving geometry, it's often helpful to first express the constraints and objective function in terms of the same variables.

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Math Problem Analysis

Mathematical Concepts

Geometry
Optimization
Differentiation

Formulas

Perimeter of semicircle: πr
Perimeter of rectangle: 2h + w
Area of semicircle: (1/2)πr^2
Area of rectangle: w * h
Derivative to find maximum: dA/dr = 0

Theorems

Optimization using differentiation

Suitable Grade Level

Grade 10-12